斐波那契数列若干项的和

王守恩

数串(1)——0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418,

0=0,
1=1,
2=2,
3=3,
4=1+3,
5=5,
6=1+5,
7=2+5,
8=8,
9=1+8,
10=2+8,
11=3+8,
12=1+3+8,
13=13,
14=1+13,
15=2+13,
16=3+13,
17=1+3+13,
18=5+13,
19=1+5+13,
20=2+5+13,
21=21,
22=1+21,
23=2+21,
24=3+21,
25=1+3+21,
26=5+21,
27=1+5+21,
28=2+5+21,
29=8+21,
30=1+8+21,
31=2+8+21,
32=3+8+21,
33=1+3+8+21,
34=34,
35=1+34,
36=2+34,
37=3+34,
38=1+3+34,
39=5+34,
40=1+5+34,
41=2+5+34,
42=8+34,
43=1+8+34,
44=2+8+34,
45=3+8+34,
46=1+3+8+34,
47=13+34,
48=1+13+34,
49=2+13+34,
50=3+13+34,
51=1+3+13+34,
52=5+13+34,
53=1+5+13+34,
54=2+5+13+34,
55=1,
56=1+55,
57=2+55,
58=3+55,
59=1+3+55,
60=5+55,

得到数串(2)——1, 1, 1, 1, 2, 1, 2, 2, 1, 2, 2, 2, 3, 1, 2, 2, 2, 3, 2, 3, 3, 1, 2, 2, 2, 3, 2, 3, 3, 2, 3, 3, 3, 4, 1, 2, 2, 2, 3, 2, 3, 3, 2, 3, 3, 3, 4, 2, 3, 3, 3, 4, 3, 4, 4, 1, 2, 2, 2, 3, 2, 3, 3, 2, 3, 3, 3, 4, 2, 3, 3, 3, 4, 3, 4, 4, 2, 3, 3, 3, 4, 3, 4, 4, 3, 4, 4, 4, 5,

a(1)=0=0,
a(2)=4=1+3,
a(3)=12=1+3+8,
a(4)=33=1+3+8+21,
a(5)=88=1+3+8+21+55,
a(6)=232=1+3+8+21+55+144,
a(7)=609=1+3+8+21+55+144+377,
a(8)=1596,
a(9)=4180,

得到数串(3)——0, 4, 12, 33, 88, 232, 609, 1596, 4180, 10945, 28656, 75024, 196417, ——是这串数吗？

求助：是这串数吗？

northwolves

A027941
a(n) = Fibonacci(2*n + 1) - 1.

0, 1, 4, 12, 33, 88, 232, 609, 1596, 4180, 10945, 28656, 75024, 196417, 514228, 1346268, 3524577, 9227464, 24157816, 63245985, 165580140, 433494436, 1134903169, 2971215072, 7778742048, 20365011073, 53316291172, 139583862444, 365435296161, 956722026040

Also: smallest number not writeable as the sum of fewer than n positive Fibonacci numbers. E.g., a(5)=88 because it is the smallest number that needs at least 5 Fibonacci numbers: 88 = 55 + 21 + 8 + 3 + 1. - Johan Claes, Apr 19 2005 [corrected for offset and clarification by Mike Speciner, Sep 19 2023] In general, a(n) is the sum of n positive Fibonacci numbers as a(n) = Sum_{i=1..n} A000045(2*i). See A001076 when negative Fibonacci numbers can be included in the sum. - Mike Speciner, Sep 24 2023
Except for first term, numbers a(n) that set a new record in the number of Fibonacci numbers needed to sum up to n. Position of records in sequence A007895. - Ralf Stephan, May 15 2005

王守恩

northwolves 发表于 2025-7-28 06:22
A027941
a(n) = Fibonacci(2*n + 1) - 1.

简单！丢了！换一串。

Table[Solve[{(Sin[Pi/6] + Sin[Pi/3] + Sin[Pi/2])^(2 n - 1) ==A*Sin[Pi/6]^(2 n - 1) + B*Sin[Pi/3]^(2 n - 1) + Sin[Pi/2]^(2 n - 1)}, {A, B}, PositiveIntegers], {n, 9}]

n=1, A(1)=1, B(1)=1,
n=2, A(2)=46, B(2)=10,
n=3, A(3)=1156, B(3)=76,
n=4, A(4)=26440, B(4)=568,
n=5, A(5)=594352, B(5)=4240,
n=6, A(6)=13318240, B(6)=31648,
n=7, A(7)=298263616, B(7)=236224,
n=8, A(8)=6678960256, B(8)=1763200,
n=9, A(9)=149557916416, B(9)=13160704,

B(n)=1, 10, 76, 568, 4240, 31648, 236224, 1763200, 13160704, 98232832, 733219840, 5472827392, 40849739776, 304906608640, 2275853910016, ——A107903——Generalized NSW numbers.——条文没我们的有意义。

A(n)=1, 46, 1156, 26440, 594352, 13318240, 298263616, 6678960256, 149557916416, 3348948866560, 74990693573632, 1679214509639680, 37601483354976256,——OEIE就没有了。

固定n, 在A(n), B(n)有解的前提下,  C(n)表示可以取到的最大值。

Table[Solve[{(Sin[Pi/6] + Sin[Pi/3] + Sin[Pi/2])^(2 n - 1) ==A*Sin[Pi/6]^(2 n - 1) + B*Sin[Pi/3]^(2 n - 1) + C*Sin[Pi/2]^(2 n - 1)}, {A, B, C}, PositiveIntegers], {n, 3}]

C(n)=1, 6, 37, 207, 1161, 6504, 36410, 203826, 1141037, 6387614, 35758350,——电脑罢工了——好不容易搞了这么几个————连我也找不到通项公式。

northwolves

$A107903(n)=\lfloor 2^{n-2}\left(1+\sqrt{3}\right)  \left(2+\sqrt{3}\right)^{n-1}\rfloor$

northwolves

A(n)=1, 46, 1156, 26440, 594352, 13318240, 298263616, 6678960256, 149557916416, 3348948866560, 74990693573632, 1679214509639680, 37601483354976256,——OEIE就没有了。
-----------------------

$A(n)=\frac{6^n}{12}  \left(\left(\sqrt{3}+3\right) \left(2-\sqrt{3}\right)^n+\left(3-\sqrt{3}\right) \left(\sqrt{3}+2\right)^n\right)-2^{2 n-1}$

northwolves

Table[Solve[{(Sin[Pi/6] + Sin[Pi/3] + Sin[Pi/2])^(2 n - 1) ==A*Sin[Pi/6]^(2 n - 1) + B*Sin[Pi/3]^(2 n - 1) + C*Sin[Pi/2]^(2 n - 1)}, {A, B, C}, PositiveIntegers], {n, 3}]

C(n)=1, 6, 37, 207, 1161, 6504, 36410, 203826, 1141037, 6387614, 35758350,——电脑罢工了——好不容易搞了这么几个————连我也找不到通项公式。
---------------------------------------
$（\frac{3}{2}+\frac{\sqrt3}{2}）^{2n-1}=A*(frac{1}{2})^{2n-1}+B*(frac{\sqrt3}{2})^{2n-1}+C}$

显然B为定值，C为左式取整: $C_n=\lfloor\frac{ \left(3+\sqrt{3}\right)^{2 n-1}}{4^n}\rfloor$